How is this a telescoping series?

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Homework Help Overview

The discussion revolves around the series $$ \sum_{k=1}^{\infty} \frac{1}{(2k-1)(2k+1)} $$ and its classification as a telescoping series. The original poster expresses confusion about the nature of the series and the expected limit as it approaches infinity.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the algebraic manipulation of the series terms, questioning the validity of splitting the fraction into separate components. There is a focus on understanding the correct form of the series and the implications of telescoping behavior.

Discussion Status

Some participants have pointed out errors in the original poster's algebraic reasoning, leading to a reevaluation of the approach. There is an ongoing exploration of how to correctly interpret the series and its components, with guidance emphasizing the importance of algebra over intuition.

Contextual Notes

The discussion highlights a misunderstanding of the algebraic manipulation of fractions, with participants questioning assumptions about the series' structure and behavior.

arhzz
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Homework Statement
Finding the limit of the series
Relevant Equations
Telescoping series formula
Hello !

Consider this series;

$$ \sum_{k=1}^{\infty} \frac{1}{(2k-1)(2k+1)} $$ It is said to find the limit of the series when approaches infinity.Now it is said that this is a telescopic series and that the limit is ##\frac{1}{2}## but I don't see it. I've split the an part (I don't know how to call it in english but it is the term within the sum symbol) I did that like this.

$$ \frac{1}{(2k-1)}+ \frac{1}{(2k+1)} $$ Now if we input k to let's say 3 we get this

$$ \frac{1}{2} + \frac{1}{3} +\frac{1}{3} + \frac{1}{5} +\frac{1}{7} + \frac{1}{7} $$. I think its fair to assume that only 1/2 should remain and the others should cancel out,that is kind of the thing by the telescoping series.But I have + across the board and the fractions won't cancel out but rather add up.I am not sure how the - is supposed to come into play here,since as I've been taught yestarteday we have to sum up the elements.The only logical solution would be that within the parantheses one element has to be negative,but I don't see how we get to that.
 
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arhzz said:
$$ \sum_{k=1}^{\infty} \frac{1}{(2k-1)(2k+1)} $$ It is said to find the limit of the series when approaches infinity.Now it is said that this is a telescopic series and that the limit is ##\frac{1}{2}## but I don't see it. I've split the an part (I don't know how to call it in english but it is the term within the sum symbol) I did that like this.

$$ \frac{1}{(2k-1)}+ \frac{1}{(2k+1)} $$
Are you sure about that?
 
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Why do you think

$$ \frac{1}{(2k-1)(2k+1)} = \frac{1}{(2k-1)}+ \frac{1}{(2k+1)} $$ ?

Plug in a value for k. Like 2.
 
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Vanadium 50 said:
Why do you think

$$ \frac{1}{(2k-1)(2k+1)} = \frac{1}{(2k-1)}+ \frac{1}{(2k+1)} $$ ?

Plug in a value for k. Like 2.
Okay so you are definitely right,these are not the same so the algebra doesn't add up.I supposed it would have to be 1st fraction - 2nd fraction but I assumed it would be + because I don't know intuition? I saw * in the denominator and simply assumed it would be +
 
Don't use intuition. Use algebra.
 
Vanadium 50 said:
Don't use intuition. Use algebra.
Yea I guess I rushed it. It all makes sense now,thanks for pointing out the mistake.
 

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